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I’m reading Randall Munroe’s (xkcd’s) book How To, which outlines several fun ways to think about everyday tasks using physics. One chapter is titled How To Have a Pool Party and in it, he outlines how to calculate the wall thickness of a pool, so that it can hold the water. Inspired by this, I am wondering if I could calculate the opposite.

The idea: to place a class cylinder in a shallow part of the North Sea, the Dogger Bank, so that we could walk around on the bottom of the sea and look around, as if in an inverted aquarium. It would have to be around 32 m tall and have a radius of, let’s say, 5 km. The exact numbers can vary, of course.

So the area of the cylinder in contact with water would be:

$$ A = 2 \pi r h$$

With a radius of 5 km and height of 32 m this works out to $ 1005309.65 m^2 $

The force on the cylinder would be:

$$ F = \rho g \left(\frac h2\right)A $$

With a density of $ \frac{997kg}{m^3} $ for water it works out to be $ 1.572 \times 10^{11} N = 157200000 kN $

The compressive strength of glass is 1000 megapascals, that's $ 1 \times 10^9 Pa = $. (According to Wikipedia, let’s assume this is correct.)

What I don’t understand is how to calculate the thickness of the glass that is needed. I understand that a force of $ 1.572 \times 10^{11} N $ would act upon an area of $ 1005309.65 m^2 $, but I can’t seem to get any further than that.

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  • $\begingroup$ Hello! It is preferable to use MathJax (LaTeX) to display formulas. You can find a tutorial at MathJax basic tutorial and quick reference. Please edit your question accordingly. Thanks! $\endgroup$
    – jng224
    Commented Jun 2, 2021 at 9:37
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    $\begingroup$ @Jonas I’ve given it my best shot, but I've never done this before, so apologies if there are mistakes. $\endgroup$
    – Flobin
    Commented Jun 2, 2021 at 10:48

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Imagine trying to increase the radius of a circular rubber band of radius $r$ under tension $T$. If an outward radial force $F$ is applied evenly all around it would do work $ W = F \times \delta r$

This work becomes potential energy in the rubber band, so $W = T \times 2 \pi \delta r$ (as the circumference increases by a different distance), so $T=\frac{F}{2 \pi}$

So you can use your force $F$ but divide by $2 \pi$.

$T = (h \rho g)\times (2 \pi rh)/(2\pi) = h^2\rho gr$

It's the same for a compressing force, $T$ is now the compressive force in the wall of the cylinder. You said the compressive strength of glass is $1 \times 10^9$Pa and has the units of pressure.

The Area is $t \times h$ where $t$ is the thickness of the glass wall. This time the area is different, it's for the cross section of the glass that gets compressed if the glass cylinder were to have a small decrease in radius. Using the above and rearranging Pressure = Force / Area, gives

$$t = \frac{\rho g h r}{1 \times 10^9}$$

For a large $r$ of 5km it's about 1.6m

Probably best to do thicker for safety! - and have a wall thickness that increases with depth, the 1.6m is at the bottom. If you chose 5m for the radius, it's 1.6mm, but could that be trusted?..it might be ok for compressibility, but any sharp impact would cause other problems, so the calculation is valid only if the other problems aren't an issue.

If you ever build it, please post here, maybe you could sell tickets!

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  • $\begingroup$ Thank you! It took me a while to understand, but I think this is correct! $\endgroup$
    – Flobin
    Commented Jun 2, 2021 at 10:51
  • $\begingroup$ It was done in a hurry, a bit more detail will be added soon $\endgroup$ Commented Jun 2, 2021 at 13:35

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